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Question-91377




Question Number 91377 by A8;15: last updated on 30/Apr/20
Commented by Tony Lin last updated on 30/Apr/20
∫(1/( (√(sinx))))dx  =∫(1/( (√(cos(x−(π/2))))))dx  =∫(1/( (√(1−2sin^2 (((2x−π)/4))))))dx  let ((2x−π)/4)=u, (du/dx)=(1/2)  2∫(1/( (√(1−2sin^2 u))))du  =F(u∣2)+c  =F(((2x−π)/4)∣2)+c
$$\int\frac{\mathrm{1}}{\:\sqrt{{sinx}}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{{cos}\left({x}−\frac{\pi}{\mathrm{2}}\right)}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\right)}}{dx} \\ $$$${let}\:\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}={u},\:\frac{{du}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {u}}}{du} \\ $$$$={F}\left({u}\mid\mathrm{2}\right)+{c} \\ $$$$={F}\left(\frac{\mathrm{2}{x}−\pi}{\mathrm{4}}\mid\mathrm{2}\right)+{c} \\ $$

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